CODEWITHSUNDEEP
bashshellsyntax
YukiNyaa
YukiNyaa

Reputation: 136

"Too many arguments" error in shell script

Here's my script:

if [ awk '$0 ~ /Failed/ { print }' $(pwd)/unity.log ]; then
    echo "Build Failed"
    exit 1
else
    echo "Build Success"
    exit 0
fi

The gist is that I am checking the file for "Build Failed" message and exiting 1 if failed.

If what I understand is correct, awk will return blank string if there is no text in the file and some text if it's found. But it shoots syntax error : ./Scripts/build.sh: line 41: [: too many arguments

Upvotes: 1

Views: 2869

Answers (3)

codeforester
codeforester

Reputation: 43039

Remove the [], use grep:

if grep -qw Failed unity.log; then
    echo "Build Failed"
    exit 1
else
    echo "Build Success"
    exit 0
fi

Upvotes: 4

Mark Reed
Mark Reed

Reputation: 95315

You could take advantage of the default exit value being 0. Also, there's no reason to do $(pwd)/filename; just do filename.

grep -qw Failed unity.log || { echo "Build Failed"; exit 1; }
echo "Build Success"

It would be also more idiomatic to send the error message on failure to stderr (with echo >&2).

Upvotes: 2

Harry Soehalim
Harry Soehalim

Reputation: 1

try this:

if [ ! "$(grep 'Build Failed' unity.log)" ]; then
    exit 1
fi

Upvotes: -2

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